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Problem : If $\sin^2\theta = \frac{x^2+y^2+1}{2x}$ , $x$ must be

(a) $1$

(b) $-2$

(c) $-3$

(d) $ 2$

My approach :

Since $0 \leq \sin^2\theta \leq 1$

$\Rightarrow 0 \leq \frac{x^2+y^2+1}{2x} \leq 1 $

$\Rightarrow x^2 +y^2+1 -2x \leq 0$

$\Rightarrow (x-1)^2 +y^2 =0$

$\Rightarrow x =1 $

Therefore option is (a)

Is it correct? please suggest thanks.

Adi Dani
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Sachin
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    Your reasoning looks sound to me – Mufasa Oct 20 '13 at 16:23
  • One line before the end I'd rather write $;(x-1)^2+y^2\le 0;$ and since this is a sum of squares it must be $,x=1;$ (and, BTW, $;y=0;$ ) , but it is very nice. – DonAntonio Oct 20 '13 at 16:26
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    A remark: when you multiply by $x$ in an inequality, you have to be sure it is positive, if not, the sign gets reversed.Luckily, since $x^2+y^2+1>0$, from the left inequality we have $x>0$. – chubakueno Oct 20 '13 at 17:09
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    I do not think we get an numeric answer with one inequality and two unknowns. – Sawarnik Oct 23 '13 at 09:32
  • We do if the inequality is a sum of squares $\le 0$. In this case, each square must be equal to $0$. – marty cohen Apr 09 '14 at 01:53

1 Answers1

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Yes the approach is right, except for one small thing, you should not just assume x to be positive while transferring variables across inequality signs. However, in this case, in the first inequality condition, if x is negative, the condition becomes:

$x^2 + y^2 + 1 <= 0$

which is wrong

x has to be positive, so doesn't really matter.

GeeYes
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